 Hi, I'm Zor. Welcome to Unizor Education. Continuing talking about theory of probabilities, I would like to present a couple of more examples of probabilistic problems. Everything which is on this lecture is just part of the whole course of advanced mathematics presented on Unizor.com. Every lecture has notes. Many problems are presented over there. And for registered students, there are exams which you can take. Everything is free. So I do suggest you to go to the website and maybe start with reading the notes which are presenting these problems I'm going to discuss. Try to solve them yourselves. That's very, very important to even if you don't solve it, at least think about how to solve these problems. And then listen to the lecture, after which I suggest you to do again, just go through the notes, read the problems and try again to recreate the solution on your own. Alright, so let's go to this particular lecture. And we are talking about lottery. I have three different problems related to the lottery. And again, this is just an example of how the probability can be used. I'm not going to go into all the intricacies of all the different lottery games, etc. I will consider only one game, which is probably one of the first ones. And the rules are, there is a random drawing of six numbers out of 49. So out of these 49, six are winning. So there is some kind of a public process, public event, when these six numbers are chosen from 49, like balls or whatever, and they're picked from the set of all the balls from 1 to 49 using some kind of a machine or whatever. So this is the random drawing. Now, before that, people who would like to participate in the game, they are purchasing tickets where they have 49 different numbers and they mark six numbers the way how they want to. After which, when the drawing is done, they check if the numbers they choose correspond to the number which were drawn by this machine, by the drawing machine. And depending on the number of the numbers which are the same on a ticket and among these six winning numbers, well, there is a certain amount of sum which you can win. And well, if you don't get two or more numbers properly guessed, so if it's zero or one, you lose your initial investment into the ticket. So the cost of ticket is completely lost. Starting from two winning numbers, which you have properly guessed out of these six, the ticket is considered as a winning ticket and the price depends on how many numbers two or above are properly guessed. Okay, so my problem right now is what's the probability of guessing K winning numbers. So again, you buy a ticket on one ticket. We're talking about one ticket right now. So you buy a ticket, you mark six numbers. Later on, there is a drawing which basically draws six numbers using this machine, whatever. And then the probability of K numbers out of whatever six you have marked on your ticket to correspond to the winning numbers picked by the machine. All right, first of all, when we're talking about probabilities, we have to talk about two different things which are very important, which is sample space, which is all the elementary events which might happen. And then we have to talk about events or event or events which we are interested in. So this is all elementary events. And this is only certain elementary events which satisfy certain condition which we are looking for. And after that, the ratio between this and this, so if you will divide this by this, you will get the probability. That's basically a definition of probability based on the frequency theory which we addressed before. All right, so what are our all elementary events and what are events we are interested in? So let's consider the chronological order. First, you buy a ticket and you mark six numbers. That process actually fixes the six numbers. You have basically say, okay, this, this, this, these are my predictions. And they are fixed at the moment when the drawing really occurs. When the drawing occurs, there can be this outcome or that outcome, basically any combination of six numbers out of 49 can be the winning combination of numbers. So we have as many elementary events as many different combinations of six numbers out of 49 we can draw from this, from this machine, which is obviously number of combinations from 49 by six. So this is the total number of elementary events where elementary event is a particular set of six numbers chosen by the drawing process. Each of these is exactly the same as far as its chances as any other. So if there are so many elementary events, we are assuming that they're all equally chanced. So they all have exactly the same probability. And since the total probability is equal to one, the total of probability of some number coming up, some set of six numbers coming up. Obviously, the sum of these should be equal to one, which means that the probability of one single event, elementary event, is equal to one over number of combinations from 49 by six. So this defines our sample space. And all elementary events are defined this way and we have a probability of each elementary event. Now we have to consider what events actually correspond to whatever our interest is. Our interest is k elements, k numbers should correspond between whatever has been drawn and whatever we have fixed before this drawing by marking certain numbers on the ticket. All right? Okay, how can we calculate that? Marking six numbers on the ticket actually breaks the all 49 different numbers into two groups. Group number one, our six numbers, which we marked on the ticket, and group two, all others, all other numbers. So there are six numbers here and 43 numbers in the second group. So we have marked six numbers and these numbers correspond to the group. They form the group one and the other 43 numbers, which we did not mark, are in the group two. Now what actually is the event of having exactly k numbers properly guessed? It means that for elementary event, as we know it, we have chosen one elementary event, a set of six numbers picked by the drawing machine. Out of these six, k is supposed to be from group one. These are the numbers we marked in the ticket. And other, so it should be k from group one and other six minus k should be from the group two. So all we have to do right now is basically count how many different combinations we can make. We can pick six numbers in such a way that k out of these six will belong to this group and other six minus k belong to this group. And this is obviously the, so how many different combinations from six by k? This how many? This many, all right? So from this group of six numbers which we have marked, k must be among winning. So basically the number of the elementary events, number of sets of six numbers which have this property is this one. And with each of these we can have this many events which have the other six minus k winning numbers picked from the group which we have not marked. So basically the combination of these, which is a product actually because for each of these we can have all of these, this is the number of different elementary events, different sets of six numbers which basically comprise our event we are interested in when k numbers are among six in the group one and six minus k in the group of not marked. And obviously since we know the number of events which we are interested in is this number and we know the all, the number of all elementary events which we have just talked about we have to divide the number of events we are interested in by the number of total events. Actually sometimes it's presented as division of number of well good events or events we are interested in, elementary events divided by the total number or which is equivalent we can say that this is the number of events which we are interested in multiplied by the probability of each event which is exactly the same thing. Since all events are equally probable all elementary events are equally probable and this is the probability so multiplying by the probability is exactly the same as divided by the number of events and this is the answer. But now let me approach this problem slightly differently it's like a second solution which is completely equivalent to this one. Now instead of buying a ticket marking the ticket with six different numbers waiting for a drawing of the numbers which is a public event actually it's usually on the TV or something like this let's reverse the sequence of events. First we draw six numbers from this drawing machine whatever but keep it secret from this particular guy who is buying a ticket then he bought a ticket and then he marked this ticket as whatever six numbers he has chosen and then we checked which numbers are the same and which ones are different. Now what's the difference actually? Well on one hand there is absolutely no difference because instead of having the ticket as the first primary chronological event which fixes six numbers and then considering the results of the drawing as elementary event we can actually do a completely reverse logic we can say that okay we have drawn six numbers from this you know using this drawing machine fixed these numbers we just don't tell this to the guy and the guy bought tickets and he randomly filled the six numbers marked in this ticket and then we compare there is absolutely no logical difference among these two things but in the first case we consider the ticket as fixed and the drawing as random as a random event in the second approach we consider the drawing as fixing six numbers and then filling the ticket as a random event basically and the result will be also the result will be the same absolutely because there is no difference between these things this is a picking of six numbers out of 49 and this is picking of six numbers of 49 in this case we are interested in k numbers being from group one and others six minus k from group two and in that case so it's basically exactly the same thing why did I mention this second approach because in another problem it will be actually used and it will be simpler to use that approach actually there is a third approach let's consider an elementary event as picking both six numbers in the drawing machine and six numbers in the in the ticket in which case the number of different combinations would be this square because I will have I will have that many different outcomes from the drawing and that many outcomes of the marking six numbers on the ticket but again you can consider it this way so that would expand the number of elementary events but it will also expand the number of events when the numbers k numbers coincide with each other and you will get exactly the same answer so my point is that there are many approaches you can use to pick one or another sample space set of elementary events and whatever now whatever way you choose should really be I mean if it's all correct ways of course you should really get the same results all right so let's go to the second problem second problem is just slightly more difficult and it's about two tickets let's say you bought two tickets but you feel them in in such a way that there are no equal numbers which you are filling these tickets so it's 12 different numbers on on two tickets six numbers and one six numbers or another and they don't coincide with each other and our our purpose is okay our purpose is to find the probability of losing both tickets now what does it mean to lose both tickets it means that one ticket should have zero or one no more than one numbers which are corresponding to the winning six numbers and the second ticket is supposed to have zero or one no more than one numbers corresponding to the winning tickets okay and we need the probability of this the way how I suggest to approach this problem is the following now again we consider our tickets to be primary so to speak and picking two tickets and marking these tickets with six numbers each basically makes certain division of all 49 numbers into in this case three groups right group number one is group number one six numbers from first ticket group number two six numbers from second ticket and group number three uh 37 numbers not marked on either ticket right so six six and thirty seven now what kind of events we actually should consider when both tickets are losing well here are event event which i um mark this way event zero zero six it means that out from the six winning numbers drawn from the from the machine winning machine we have zero numbers in the first group zero numbers in the second group and all six are other numbers which means all six winning numbers belong to the group which we uh did not mark at all another combination is when you have zero numbers winning numbers from the first group from the first ticket one number from the second and five numbers it should be six all together right so one from this group and five from this group it means that the second ticket has only one number corresponding to the winning and this one has none now another is one zero five which means that the first group the first ticket has one winning number properly guessed this one the second ticket has zero and all other five winning numbers from this group and finally one one four when you have one number one by this one number one for this ticket and other four winning numbers belong to this group so these are four events and there are no more other events so uh any other event means that i have two or more winning numbers in one of these groups and that means that the that the corresponding ticket would win something and we are not uh looking for this so only these four events we have to really count and all we need to do is the probability of each of them right so the probability can be counted as obviously as number of events which number of elementary events which fall into this event category all right and this is what if i have zero from the first group number of combinations is c six zero number of um combinations from six by zero times from six by zero for the second group and times from 37 by 6 for the third group now this one has zero from 6 by 0 for the first group from 6 by 1 for the second group and for 37 by 5 for the second now this has exactly the same thing just change the places and the last one i can pick a number of cars number of combinations from six by one one winning number from this one winning number from this and four winning numbers from this so these are number of combinations of elementary events which make up this this this or this event so if we add them up and divide by the total number of different combinations of uh elementary events different elementary events this is our sample space size um you will get the probability of this and uh it's basically the following so this can be simplified this one is one one and this is c 37 by six this is one this is six so it's six 37 by five this is also six uh 37 by five and this is six times six 36 37 by four so some of these divided by the total number of combinations i have counted is 0.71 so 0.71 is a probability of losing two tickets if you fill them up with all different numbers in the story and the third problem is somewhat similar to the first one but just a little bit more complicated uh you have uh also you have certain number of tickets which you buy now there is no restriction on how you fill them up you completely randomly fill them up each one of those and tickets okay and tickets and we are also looking about looking for losing these tickets so obviously the probability of losing all the tickets is diminishing as the number and grows because you're increasing your chances to win something right so my question is um what's the minimum number of tickets which you have to buy to reduce your probability below 50 percent below 0.5 so we are interested in less than 0.5 probability of losing all tickets losing means that every one of those wins either 0 or 1 winning number all right so how can i approach this now let's go back to the first problem if you remember i presented two different approaches considering your ticket as primary which fixes a certain number of uh numbers and then the the random process is uh drawing the winning numbers from this machine or alternatively you can draw the winning numbers first just don't tell it to the guy so he buys the ticket and feeling the ticket is basically the the random process which is completely equivalent to the previous one so in this case we will use the second approach because we are buying more the more than one tickets so it's not like two independent things um it's one drawing and and tickets so it's more convenient to have one drawing first but kept secret so it fixes a certain number of winning numbers so we have six fixed numbers they are the winners all right now we are buying and tickets one after another filling up randomly with whatever numbers we want and basically compare what what what's the uh result is and what what kind of elementary events would constitute the event that none of these is winning anything all right so this is easier because this gives us the chance to basically uh multiply the number of elementary events and here is why for one particular for one particular ticket well first of all let's talk about elementary events in this particular case since i'm buying and tickets and each ticket i fill up with six uh numbers so n n sets of uh let's call it strings of six numbers is an elementary event so my first ticket for instance gives me some numbers one number two number three four five and six this is my first ticket my first part of one long elementary event then we buy another ticket and we have another six numbers etc and then we have the n's ticket and n strings and each strings have each string has six numbers from from one to two to 49 each uh order actually is uh unimportant but all together this n strings where each string contains six numbers basically chosen from one to 49 uh is one elementary event okay that that's that's what my purchasing and tickets and filling them up randomly means all right now if this is an elementary event all we have to do is find out the number of elementary event when the first ticket is losing and the second ticket is losing and the n's ticket is losing all right which means that among these there are only zero or one common numbers with my six winning so i have six winning before the drawing was done right so in this i have to have no more than one so zero or one and we actually know it's um number of combinations from six uh by zero times number of combinations from four to three by six that's zero number of combinations in one particular string only plus by one and that would be this so either zero winning and all six belong to non-chosen amount here or one winning and five belongs to all others four to three so the sum of these is basically the number of different strings of the lengths six which are not really having more than one common numbers but now this also has the same number of losing strings right so this is a number of losing strings number one and this is a number of losing strings number two which is exactly the same and losing strings number n so i have to raise this to the n's degree because with each of these each of these each of these etc these are all different combinations and to find out the total of uh n strings uh with each of them being having you know this particular number of losing strings i have to multiply these all losing strings with every losing this string i have every losing this string and every losing that string so this is the number of elementary events which we are interested in and obviously the number of all the combinations when um we when we uh have uh all the different combinations of n strings with six numbers each well it's basically same thing again it's this to the n's degree because each string has that many different combinations right it's six out of 49 and if with each of these i have each of these with each of these and that's why i multiply them all together so if you will divide this by this you will get the answer to the problem which basically means that you have to have one ticket and and raise it to the to the power of n well that's basically it that that's my end of the lottery games i hope you are old winners and i'm sure actually you you you are if you if you understand whatever all these problems are about i do suggest you to read again the notes of this lecture and try again to recreate the solution it's written relatively in in details in in the notes for the lecture so you're welcome to go to unizord.com and review this lecture again it's very useful that's it thank you very much and good luck